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Quoting from Wikipedia,

The common-ion effect refers to the decrease in solubility of an ionic precipitate by the addition to the solution of a soluble compound with an ion in common with the precipitate. This behaviour is a consequence of Le Chatelier's principle for the equilibrium reaction of the ionic association/dissociation.

Let us consider an example of $\ce{AgCl}$ in aqueous solution of $\ce{NaCl}$. The following equilibrium exsist in the solution, $$ \ce{ AgCl(s) <<=> Ag+(aq) + Cl-(aq)} $$ According to the common-ion effect, as we increase the concentration of $\ce{NaCl}$ in the solution the equilibrium should shifts towards left side i.e., solubility of $\ce{AgCl}$ decreases with increase in the concentration of $\ce{Cl-}$ ions. But what I think here is that, can't the excess $\ce{Cl-}$ ions forms a soluble complex with the $\ce{AgCl}$ precipitate? That reaction is as follows, $$ \ce{ AgCl + Cl- <=> [AgCl2]-} $$ If that's the case then, how to judge the solubility with these two competing factors?

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    $\begingroup$ If by 'judge' you mean 'calculate', then the answer is: make a system with all the equilibrium expressions involved, meaning the solubility product, equilibrium constant for the complex, and whatever else is needed, and solve. This is actually a rather frequent situation in inorganic analytical chemistry. You can stop certain rather insoluble salts from precipitating, by forming complexes. The effect of pH on precipitation is a related example. Just take care of handling the case where the precipitate disappears, because then you no longer have the solubility product. $\endgroup$ Aug 24, 2022 at 18:22
  • $\begingroup$ BTW, just remembered that this happens with silver and cyanide: en.wikipedia.org/wiki/Silver_cyanide#Reactions . As you add more cyanide, the initially formed precipitate redissolves. $\endgroup$ Aug 24, 2022 at 18:25
  • $\begingroup$ If there is some serious complexation (high K) than common ion is irrelevant. $\endgroup$
    – Mithoron
    Aug 24, 2022 at 19:16
  • $\begingroup$ OTOH, there can be precipitation of Ag+ and AgX2^-, or Hg^2+ and [HgX4]^2-, if the excess of the complexant was just temporary. E.g. during progressive addition of metal salt solution to the solution of the complexant. $\endgroup$
    – Poutnik
    Aug 26, 2022 at 13:06

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If that's the case then, how to judge the solubility with these two competing factors?

Solubility is how much of a single substance you can dissolve in pure solvent. For the common ion effect, you add a second solute, so at that point it does not make sense to talk about solubility anymore. Thus, the Wikipedia definition cited is a little bit awkward and could be improved. Here is another definition that avoids the problem:

The common-ion effect is used to describe the effect on an existing equilibrium by the addition of a second substance that contains an ion common to the equilibrium. If several salts are present in a system, they all ionize in the solution.

Source: https://ecampusontario.pressbooks.pub/genchemforgeegees/chapter/6-1-common-ion-effect/

Instead, you can discuss how much solid precipitated, as an easy quantitative description of what is going on.

If you want to demonstrate the common ion effect for the reaction $$ \ce{ AgCl(s) <<=> Ag+(aq) + Cl-(aq)} $$ you can add some silver nitrate. This will increase the mass of precipitated silver chloride as predicted from the common ion effect.

Apart from complexation, precipitation of other species also lead to results you might not expect at first. If you add sodium bromide (no common ion), it will precipitate some silver bromide, and some silver chloride will dissolve.

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  • $\begingroup$ @Poutnik I think a simple and limited answer is a first step. A second answer delving into the complexity of complexation would be great, and I invite everyone (except me) to write it. $\endgroup$
    – Karsten
    Aug 26, 2022 at 12:37
  • $\begingroup$ Sure. But we need not extremes. :-) $\endgroup$
    – Poutnik
    Aug 26, 2022 at 12:45

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